某学生在整理班级同学的课外阅读情况时,随机抽取了20名学生进行调查,记录他们每周课外阅读的小时数如下:3, 5, 4, 6, 3, 7, 5, 4, 6, 5, 3, 4, 6, 7, 5, 4, 3, 5, 6, 4。如果该学生想用一个统计量来代表这组数据的集中趋势,并且希望这个值能反映大多数学生的阅读时间,那么他应该选择以下哪个统计量?
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首先,求原抛物线 y = x² - 4x + 3 的顶点:配方得 y = (x - 2)² - 1,顶点为 (2, -1)。点 P 在第一象限且在抛物线上,且到 x 轴距离为 1,即纵坐标为 1。代入抛物线方程:1 = x² - 4x + 3,解得 x² - 4x + 2 = 0,解得 x = 2 ± √2。因在第一象限,取 x = 2 + √2,故 P(2 + √2, 1)。又 P 在反比例函数 y = k/x 上,代入得 k = x·y = (2 + √2)·1 = 2 + √2,故反比例函数为 y = (2 + √2)/x。将原抛物线绕顶点 (2, -1) 旋转 180°,其开口方向反向,形状不变,新抛物线方程为 y = -(x - 2)² - 1 = -x² + 4x - 5。联立新抛物线与反比例函数:-x² + 4x - 5 = (2 + √2)/x,两边乘以 x(x ≠ 0)得:-x³ + 4x² - 5x = 2 + √2,即 -x³ + 4x² - 5x - (2 + √2) = 0。此三次方程在实数范围内分析图像趋势:当 x → 0⁺ 时,左边 → -∞;当 x → +∞ 时,-x³ 主导,→ -∞;在 x = 2 附近函数值变化分析可知,函数图像仅穿过 x 轴一次,故仅有一个实数解。因此,新抛物线与反比例函数图像有 1 个交点。
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[{"id":478,"content":"周一","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"待完善","explanation":"解析待完善","options":[]},{"id":193,"content":"小明买了3支铅笔和2本笔记本,共花费18元。已知每支铅笔2元,那么每本笔记本多少钱?","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"A","explanation":"首先计算3支铅笔的总价:3 × 2 = 6(元)。小明一共花了18元,因此2本笔记本的总价为:18 - 6 = 12(元)。那么每本笔记本的价格为:12 ÷ 2 = 6(元)。所以正确答案是A。","options":[{"id":"A","content":"6元"},{"id":"B","content":"5元"},{"id":"C","content":"4元"},{"id":"D","content":"3元"}]},{"id":2235,"content":"某学生在数轴上从原点出发,先向右移动5个单位长度,再向左移动8个单位长度,接着向右移动3个单位长度,最后向左移动4个单位长度。此时该学生所在位置的数与其相反数之和为___。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"困难","answer":"0","explanation":"首先计算该学生在数轴上的最终位置:从原点0开始,向右移动5个单位到达+5,再向左移动8个单位到达-3,接着向右移动3个单位到达0,最后向左移动4个单位到达-4。因此,最终位置表示的数是-4。一个数与其相反数之和恒为0,即-4 + 4 = 0。本题综合考查了数轴上的正负数移动、有理数加减运算以及相反数的性质,符合七年级正负数章节的拓展要求,难度较高。","options":[]},{"id":1095,"content":"在一次班级数学测验中,某学生记录了五名同学的数学成绩(单位:分)分别为:85,92,78,90,85。这组数据的众数是____。","type":"填空题","subject":"数学","grade":"七年级","stage":"小学","difficulty":"简单","answer":"85","explanation":"众数是一组数据中出现次数最多的数。在这组数据中,85出现了两次,而其他分数(92、78、90)各出现一次,因此众数是85。","options":[]},{"id":2156,"content":"某学生在数轴上标记了三个有理数:-1.5、0.8 和 -2\/3。若将这三个数按从小到大的顺序排列,正确的结果是?","type":"选择题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"D","explanation":"首先比较负数:-1.5 比 -2\/3(约等于 -0.67)更小,因为它在数轴上更靠左;0.8 是正数,最大。因此从小到大的顺序是 -1.5 < -2\/3 < 0.8。选项 D 正确。","options":[{"id":"A","content":"-1.5, -2\/3, 0.8"},{"id":"B","content":"-2\/3, -1.5, 0.8"},{"id":"C","content":"0.8, -2\/3, -1.5"},{"id":"D","content":"-1.5, -2\/3, 0.8"}]},{"id":668,"content":"在一次班级环保活动中,某学生记录了5天内每天收集的废纸重量(单位:千克):3,5,4,6,2。为了估算一个月(按30天计算)的废纸收集总量,他先求出这5天的平均每天收集量,再乘以30。那么,他计算出的月收集总量是___千克。","type":"填空题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"120","explanation":"首先计算5天收集废纸的平均重量:(3 + 5 + 4 + 6 + 2) ÷ 5 = 20 ÷ 5 = 4(千克\/天)。然后用平均每天收集量乘以30天:4 × 30 = 120(千克)。因此,估算的月收集总量是120千克。本题考查数据的收集与整理中的平均数计算及其应用,属于简单难度的实际问题建模。","options":[]},{"id":620,"content":"72度","type":"选择题","subject":"数学","grade":"初一","stage":"小学","difficulty":"简单","answer":"待完善","explanation":"解析待完善","options":[]},{"id":2215,"content":"某学生在记录一周内每天的温度变化时,发现某天的气温比前一天上升了5℃,记作+5℃;而另一天的气温比前一天下降了3℃,应记作____℃。","type":"填空题","subject":"数学","grade":"七年级","stage":"初中","difficulty":"简单","answer":"-3","explanation":"根据正数和负数表示相反意义的量的规则,气温上升用正数表示,下降则用负数表示。因此,气温下降3℃应记作-3℃。此题考查学生对正负数在实际情境中应用的理解,符合七年级正负数表示相反意义的量的知识点。","options":[]},{"id":299,"content":"某学生在平面直角坐标系中画了一个点,该点的横坐标是-3,纵坐标是5。这个点位于第几象限?","type":"选择题","subject":"数学","grade":"初一","stage":"初中","difficulty":"简单","answer":"B","explanation":"在平面直角坐标系中,四个象限的划分如下:第一象限横纵坐标均为正,第二象限横坐标为负、纵坐标为正,第三象限横纵坐标均为负,第四象限横坐标为正、纵坐标为负。题目中给出的点横坐标是-3(负),纵坐标是5(正),因此该点位于第二象限。","options":[{"id":"A","content":"第一象限"},{"id":"B","content":"第二象限"},{"id":"C","content":"第三象限"},{"id":"D","content":"第四象限"}]},{"id":266,"content":"某学生在解方程 3(x - 4) = 2x + 5 时,第一步将等式两边同时展开,得到 3x - 12 = 2x + 5。接下来,他将含 x 的项移到等式左边,常数项移到右边,得到 ___ = ___。","type":"填空题","subject":"数学","grade":"初一","stage":"小学","difficulty":"中等","answer":"3x - 2x = 5 + 12","explanation":"根据解一元一次方程的步骤,移项时要改变项的符号。原式为 3x - 12 = 2x + 5。将 2x 移到左边变为 -2x,将 -12 移到右边变为 +12,因此得到 3x - 2x = 5 + 12。这是移项法则的正确应用,体现了等式两边同时加减同一个整式的变形规则。","options":[]}]